Question 7 of 11.5 step 2 of 2
$13 / 23$
Correct
A random sample of 9 fields of corn has a mean yield of 34.7 bushels per acre and standard deviation of 8.04 bushels per acre. Determine the $98 \%$ confidence interval for the true mean yield. Assume the population is approximately normal.

Step 2 of 2 : Construct the $98 \%$ confidence interval. Round your answer to one decimal place.
Answer
How to enter your answer (opens in new window)
Tables
Keyboard Sh

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 34.7 bushels per acre, the Z-score (\(Z_{\alpha/2}\)) for a 98% confidence interval is approximately 2.33, the standard deviation (\(\sigma\)) is 8.04 bushels per acre, and the sample size (\(n\)) is 9.

Step 2 ：We use the formula for the confidence interval which is given by: \[ \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \]

Step 3 ：Substitute the given values into the formula to calculate the confidence interval: \[34.7 \pm 2.33 \times \frac{8.04}{\sqrt{9}}\]

Step 4 ：The margin of error is calculated to be approximately 6.2444.

Step 5 ：The lower limit of the confidence interval is calculated as \(34.7 - 6.2444 = 28.5\)

Step 6 ：The upper limit of the confidence interval is calculated as \(34.7 + 6.2444 = 40.9\)

Step 7 ：The $98 \%$ confidence interval for the true mean yield is \(\boxed{(28.5, 40.9)}\) bushels per acre.